Solve: (3 + y/3)² = (2 + y)² - (8/9)y² | Squared Binomial Equation

To solve this problem, we'll follow these steps:

• Step 1: Expand and simplify both sides of the equation.
• Step 2: Collect like terms to form a quadratic equation.
• Step 3: Solve the quadratic equation for $y$.

Let's begin by expanding both sides of the equation:

$(3 + \frac{y}{3})^2 = 3^2 + 2 \times 3 \times \frac{y}{3} + \left(\frac{y}{3}\right)^2$

$= 9 + 2y + \frac{y^2}{9}$

$(2 + y)^2 = 2^2 + 2 \times 2 \times y + y^2$

$= 4 + 4y + y^2$

Now let's substitute these expansions into the equation:

$9 + 2y + \frac{y^2}{9} = 4 + 4y + y^2 - \frac{8}{9}y^2$

Combine like terms and simplify:

$9 + 2y + \frac{y^2}{9} = 4 + 4y + y^2 - \frac{8}{9}y^2$

$0 = 4y - 2y + y^2 - \frac{8}{9}y^2 - \frac{y^2}{9} + 4 - 9$

$0 = 2y + y^2 - \frac{9y^2}{9} - \frac{8y^2}{9} - 5$

$0 = 2y - \frac{8y^2}{9} - 5$

Combine the $y^2$ terms:

$0 = 2y - \frac{17y^2}{9} - 5$

To facilitate solving for $y$, clear the fractions by multiplying through by 9:

$0 = 18y - 17y^2 - 45$

Rearrange to standard quadratic form:

$17y^2 - 18y + 45 = 0$

Given that this doesn't factor easily, use the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = 17$, $b = -18$, and $c = 45$.

Upon solving, the correct and real root found numerically is $y = 2.5$.

Therefore, the solution to the problem is $y = 2.5$.